Results 1 to 4 of 4

Thread: Finding the Line Integral

  1. #1
    Super Member
    Joined
    Dec 2008
    Posts
    509

    Finding the Line Integral

    Hi
    I am having trouble on working out the following question:

    Given C is the anticlockwise circular path $\displaystyle x^2+y^2=4$, starting and ending at (0,2) evaluate the line integrals $\displaystyle \oint_C F.dr$ where $\displaystyle F = xyi+xyj$

    This is what i have done

    $\displaystyle r=xi + yj$
    $\displaystyle dr = i + j$

    $\displaystyle x=2cos(\theta)$, $\displaystyle y=2sin(\theta)$

    $\displaystyle F = (4cos(\theta) sin(\theta))i + (4cos\theta sin\theta )j$

    $\displaystyle \int_{0}^(2\pi) (4cos\theta sin\theta)i + (4cos\theta sin\theta )j \cdot i + j$

    P.S
    Last edited by Paymemoney; May 27th 2011 at 05:03 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Dec 2010
    Posts
    18
    $\displaystyle dr\neq \hat{\imath} + \hat{\jmath}$
    $\displaystyle dr = dx\hat{\imath} + dy \hat{\jmath}$

    If
    $\displaystyle x=2\cos\theta$, $\displaystyle y=2\sin\theta$

    what are dx and dy?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Dec 2008
    Posts
    509
    ok i have attempted this question again and this is what i have done but this is incorrect.


    This is what i have have done:
    $\displaystyle \int_0^{2\pi} 2cos(\theta)+2cos(\theta)2sin(\theta)*2cos(\theta)-2sin(\theta) d(\theta)$

    $\displaystyle \int_0^{2\pi} 8cos(\theta)^2 * sin(\theta) - 8cos(\theta) * sin(\theta)^2 + 8cos(\theta)^2 * sin(\theta) - 8cos(\theta)* sin(\theta)^2$

    $\displaystyle \int_0^{2\pi} 8(cos(\theta)^2 * sin(\theta) - cos(\theta) * sin(\theta)^2 + cos(\theta)^2 * sin(\theta) - cos(\theta)* sin(\theta)^2)$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,769
    Thanks
    3027
    Without parentheses, it is almost impossible to read what you have written. With $\displaystyle x= 2 cos(\theta)$, $\displaystyle y= 2 sin(\theta)$, $\displaystyle dx= -2 sin(\theta)dt$ and $\displaystyle dy= 2 cos(\theta)dt$. F(x,y) becomes $\displaystyle 4sin(\theta)cos(\theta)i+ 4 sin(\theta)cos(\theta)j$ and so the integral is
    $\displaystyle 8\int_{\pi/2}^{5\pi/2} (-sin(\theta)cos^2(\theta)+ sin^2(\theta)cos(\theta))dt$
    Personally, I would have used $\displaystyle x= 2sin(\theta)$, $\displaystyle y= 2cos(\theta)$ so I could integrate from 0 to $\displaystyle 2\pi$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Dec 11th 2011, 11:30 PM
  2. [SOLVED] Line Integral along a straight line.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Apr 11th 2011, 07:18 PM
  3. Replies: 1
    Last Post: Sep 20th 2010, 11:50 AM
  4. Finding parametrizattion for a Line Integral
    Posted in the Calculus Forum
    Replies: 6
    Last Post: May 29th 2010, 10:47 AM
  5. Finding a line perpendicular to another line
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: Dec 12th 2009, 09:13 PM

Search Tags


/mathhelpforum @mathhelpforum